Question Number 20337 by Tinkutara last updated on 25/Aug/17 $$\mathrm{Why}\:\mathrm{oxidising}\:\mathrm{character}\:\mathrm{of}\:\mathrm{F}_{\mathrm{2}} \:>\:\mathrm{Cl}_{\mathrm{2}} ? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20334 by Tinkutara last updated on 25/Aug/17 $$\mathrm{Choose}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{regarding}\:\mathrm{E}.\mathrm{N}. \\ $$$$\left(\mathrm{1}\right)\:\mathrm{B}\:>\:\mathrm{Al}\:>\:\mathrm{Ga}\:>\:\mathrm{In} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{B}\:>\:\mathrm{Al}\:=\:\mathrm{Ga}\:=\:\mathrm{In} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{B}\:>\:\mathrm{In}\:>\:\mathrm{Ga}\:=\:\mathrm{Al} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{B}\:>\:\mathrm{In}\:>\:\mathrm{Ga}\:>\:\mathrm{Al} \\ $$ Terms of Service Privacy Policy…
Question Number 85858 by jagoll last updated on 25/Mar/20 $$\mathrm{Is}\:\mathrm{a}\:\mathrm{matrix} \\ $$$$\mathrm{A}^{\mathrm{T}} \mathrm{A}\:\mathrm{always}\:\mathrm{positive}\:\mathrm{definite}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20316 by Tinkutara last updated on 25/Aug/17 $$\mathrm{If}\:\mathrm{at}\:\mathrm{a}\:\mathrm{height}\:\mathrm{of}\:\mathrm{40}\:\mathrm{m},\:\mathrm{the}\:\mathrm{direction}\:\mathrm{of} \\ $$$$\mathrm{motion}\:\mathrm{of}\:\mathrm{a}\:\mathrm{projectile}\:\mathrm{makes}\:\mathrm{an}\:\mathrm{angle} \\ $$$$\pi/\mathrm{4}\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal},\:\mathrm{then}\:\mathrm{its}\:\mathrm{initial} \\ $$$$\mathrm{velocity}\:\mathrm{and}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{projection}\:\mathrm{are}, \\ $$$$\mathrm{respectively} \\ $$$$\left({a}\right)\:\mathrm{30},\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(−\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\left({b}\right)\:\mathrm{30},\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\…
Question Number 20200 by ketto last updated on 24/Aug/17 Answered by Joel577 last updated on 24/Aug/17 $${x}\:.\:\mathrm{log}\:\mathrm{2}\:=\:\mathrm{6}{x}\:+\:\mathrm{3} \\ $$$$\left({x}\:.\:\mathrm{log}\:\mathrm{2}\right)\:−\:\mathrm{6}{x}\:=\:\mathrm{3} \\ $$$${x}\left[\left(\mathrm{log}\:\mathrm{2}\right)\:−\:\mathrm{6}\right]\:=\:\mathrm{3} \\ $$$${x}\:=\:\frac{\mathrm{3}}{\left(\mathrm{log}\:\mathrm{2}\right)\:−\:\mathrm{6}} \\ $$…
Question Number 20182 by DKumar last updated on 23/Aug/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20177 by DKumar last updated on 23/Aug/17 $${But}\:{ans}\:{is}\:\frac{\mathrm{1}}{\mathrm{108}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20174 by DKumar last updated on 23/Aug/17 $$\left(\mathrm{0}.\overset{−} {\mathrm{1}}\right)^{\mathrm{2}} \left\{\mathrm{1}−\mathrm{9}\left(\mathrm{0}.\mathrm{1}\overset{−} {\mathrm{6}}\right)^{\mathrm{2}} \right\} \\ $$ Answered by mrW1 last updated on 23/Aug/17 $$\mathrm{0}.\overline {\mathrm{1}}=\mathrm{0}.\mathrm{111111}…=\frac{\mathrm{1}}{\mathrm{9}}…
Question Number 85698 by Rio Michael last updated on 24/Mar/20 $$\mathrm{please}\:\mathrm{any}\:\mathrm{recommendation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{youtube}\:\mathrm{video} \\ $$$$\mathrm{on}\:\mathrm{General}\:\mathrm{conics}?? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 85676 by john santu last updated on 24/Mar/20 $$\int\underset{\mathrm{0}} {\overset{\infty} {\:}}\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$${let}\:{x}\:=\:\mathrm{tan}\:{t}\:\Rightarrow{dx}=\mathrm{sec}\:^{\mathrm{2}} {t}\:{dt} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} {t}\:{dt}}{\left(\mathrm{tan}\:{t}+\mathrm{sec}\:{t}\right)^{\mathrm{2}} }\:=\: \\…